SECTION 1– DESIGN FOR SIMPLE STRESSES
TENSION, COMPRESSION, SHEAR
DESIGN PROBLEMS
1. The link shown, made of AISI C1045 steel, as rolled, is subjected to a tensile load of
8000 lb. Let h = 5.1 b . If the load is repeated but not reversed, determine the
dimensions of the section with the design based on (a) ultimate strength, (b) yield
strength. (c) If this link, which is 15 in. long., must not elongate more than 0.005
in., what should be the dimensions of the cross section?
Problems 1 – 3.
Solution:
For AISI C1045 steel, as rolled (Table AT 7)
s ksi
u = 96
s ksi y = 59
E psi 6
= 30⋅10
FA
sd =
where
F = 8000 lb
A = bh
but
h = 5.1 b
therefore 2 A = 5.1 b
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
FA
s
u
s
d= = N
96,000 =
8000
1.5
b
6
5
2
.
b = .0 577 in say in
8
2
SECTION 1– DESIGN FOR SIMPLE STRESSES
15
= 5.1 =
h b in
16
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
FA
s
u
s
d= = N
59,000 =
8000
1.5
b
3
9
2
.
b = .0 521in say in 16
27
h b in
= 5.1 =
32
FL δ =
(c) Elongation = AE
where,
δ = .0 005 in
F = 8000 lb
E psi 6
= 30⋅10
L =15 in
2
A = 5.1 b
then,
FL δ =
AE
( )( )
8000 15 .0 005
⋅
=
b
( )( ) 2 6
5.1 30 10
3
.
b = .0 730 in say in 4
1
h b in
= 5.1 8
=1
2. The same as 1 except that the material is malleable iron, ASTM A47-52, grade 35 018.
Solution:
For malleable iron, ASTM A47-52, grade 35 018(Table AT 6)
s ksi
u = 55
s ksi y = 36 5.
E psi 6
= 25⋅10
3
SECTION 1– DESIGN FOR SIMPLE STRESSES
FA
sd =
where
F = 8000 lb
A = bh
but
h = 5.1 b
therefore 2 A = 5.1 b
(a) Based on ultimate strength
N = factor of safety = 6 for repeated but not reversed load (Table 1.1)
FA
s
su
d= = N
2
55,000 6
=
7
1.5
.
8000 b
b = .0 763 in say in 8
5
h b in
= 5.1 16
=1
(b) Based on yield strength
N = factor of safety = 3 for repeated but not reversed load (Table 1.1)
FA
s
u
s
d= = N
2
36,500 3
=
11
1.5
.
8000 b
b = .0 622 in say in
16
1
h b in
= 5.1 =1
32
FL δ =
(c) Elongation = AE
where,
δ = .0 005 in
F = 8000 lb
E psi 6
= 25⋅10
L =15 in
2
A = 5.1 b
then,
4
SECTION 1– DESIGN FOR SIMPLE STRESSES
FL δ = AE( )( )
8000 15 .0 005
⋅
=
b
( )( ) 2 6
5.1 25 10
7
.
b = 8.0 in say in 8
5
h b in
= 5.1 16
=1
3. The same as 1 except that the material is gray iron, ASTM 30.
Solution:
For ASTM 30 (Table AT 6)
s ksi
u = 30 , no y
s
E psi 6
=14 5. ⋅10
Note: since there is no y
s for brittle materials. Solve only for (a) and (c)
FA
sd =
where
F = 8000 lb
A = bh
but
h = 5.1 b
therefore 2 A = 5.1 b
(a) Based on ultimate strength
N = factor of safety = 7 ~ 8 say 7.5 (Table 1.1)
FA
s
su
d= = N
30,000 =
8000
1.5
b
7.5
3
2
b = .1 1547 in say in
1.
16
25
h b in
= 5.1 =1
32
FL δ =
(c) Elongation = AE
where,
δ = .0 005 in
F = 8000 lb
E psi 6
=14 5. ⋅10
5
SECTION 1– DESIGN FOR SIMPLE STRESSES
L =15 in
A = 5.1
b then,
2
FL δ =
( )( )
AE
8000 15 .0 005 =
⋅ b
( )( ) 2 6
5.1 14 5. 10
1
b = .1 050 in say in
1.
16
19
h b in
= 5.1 32
=1
4. A piston rod, made of AISI 3140 steel, OQT 1000 F (Fig. AF 2), is subjected to a
repeated, reversed load. The rod is for a 20-in. air compressor, where the
maximum pressure is 125 psig. Compute the diameter of the rod using a design
factor based on (a) ultimate strength, (b) yield strength.
Solution:
From Fig. AF 2 for AISI 3140, OQT 1000 F
s ksi
u =152 5.
s ksi y =132 5.
π
F force ( ) ( ) 20 125 39,270 lb 39 27. kips
====
2
4
From Table 1.1, page 20
= 8 Nu
Ny = 4
(a) Based on ultimate strength
NFu
A=
39 27.
2
u
π
s
( )( ) 8
152.5 5
1
4
d=
d = 62.1 in say in 8
(b) Based on yield strength
NF
A=
y
y
4
( )( ) 4 d =
π
s
39 27. 132.5
2
6
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
1
d = 23.1 in say in
4
5. A hollow, short compression member, of normalized cast steel (ASTM A27-58, 65
ksi), is to support a load of 1500 kips with a factor of safety of 8 based on the
ultimate strength. Determine the outside and inside diameters if Do Di = 2 .
Solution:
u = 65
s ksi
= 8 Nu
F =1500 kips
2
πππ
3
( ) ( )4
2222
ADDDD
= − = − = oiii
44
2
DNF
3
i
4
D
π
A
= = = iu
( )( ) 8 1500
4
u
s
65
7
D in i = 85.8 say in


Do Diin = =
8
7
2 2 8  17
=
8


8
3
4
6. A short compression member with Do Di = 2 is to support a dead load of 25 tons. The
material is to be 4130 steel, WQT 1100 F. Calculate the outside and inside
diameters on the basis of (a) yield strength, (b) ultimate strength.
Solution:
From Table AT 7 for 4130, WQT 1100 F
s ksi
u =127
s ksi y =114
From Table 1.1 page 20, for dead load
= 3 ~ 4 Nu, say 4
Ny = 5.1 ~ 2 , say 2
3
πππ
Area, (
) ( )4
2222
2
D
i
ADDDD
=−=−=
4
oiii
44
F = 25 tons = 50 kips
(a) Based on yield strength
2
DNF3
Aπ
===
iy
( )( )
2 50
4
s
y
114
7
SECTION 1– DESIGN FOR SIMPLE STRESSES
5
D in i = 61.0 say in
8
Do Diin = =


22=

8
(b) Based on ultimate strength
= = = iu
32
DNF
( )( ) 4 50
π
A
4

5
s
u
D in i = 82.0 say in
8
Do Diin = =


22=
8
7

1
1
4
7
127

3
1
4
7. A round, steel tension member, 55 in. long, is subjected to a maximum load of 7000 lb.
(a) What should be its diameter if the total elongation is not to exceed 0.030 in?
(b) Choose a steel that would be suitable on the basis of yield strength if the load
is gradually applied and repeated (not reversed).
Solution:
FL
δ = or E
FL A
(a) AE
δ
=
where,
F = 7000 lb
L = 55 in
δ = .0 030
in E psi 6
= 30⋅10
( )( )
π
7000 55
2
A=d=
( )( ) 6
4⋅
.0 030 30 10
3
d = 74.0 in say in
4
(b) For gradually applied and repeated (not reversed) load
Ny = 3
7000
NF
y
s
( )( ) 3
y
47,534 = = =
psi
A
π
()
2
75.0
4
s ksi y ≈ 48
say C1015 normalized condition (s ksi y = 48 )
8. A centrifuge has a small bucket, weighing 0.332 lb. with contents, suspended on a
manganese bronze pin (B138-A, ½ hard) at the end of a horizontal arm. If the pin
is in double shear under the action of the centrifugal force, determine the diameter
8
SECTION 1– DESIGN FOR SIMPLE STRESSES
needed for 10,000 rpm of the arm. The center of gravity of the bucket is 12 in.
from the axis of rotation.
Solution:
From Table AT 3, for B138-A, ½ hard
s ksi
us = 48
W 2
F
=ω
g
where r
W = .0 332 lb
2
g = 32 2. fps
( )1047 sec
ω
2
rad n
ππ
60
60
r =12 in
W
2 10,000
===
F 1047 1 11,300 11 3.
2.0 332 2
r ( ) ( ) lb kips = ω = = =
g
32 2.
From Table 1.1, page 20
N = 3 ~ 4 , say 4
u
A=
( )( )
N F u 4 11 3.
s
π
for double shear
  
 d
22 =
d = .0 774 in say in
32
4

48
25
CHECK PROBLEMS
3
9. The link shown is made of AISIC1020 annealed steel, with b in
= and
4
1
=1 . (a) What force will cause breakage? (b) For a design factor of 4 based
h in
2
on the ultimate strength, what is the maximum allowable load? (c) If N = 5.2
based on the yield strength, what is the allowable load?
Problem 9.
9
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
For AISI C1020 annealed steel, from Table AT 7
s ksi
u = 57
s ksi y = 42
(a) F = su A
3
1
1
A bh 
= in

==


4
2

2
  
.1 125

F = (57)( .1 125) = 64 kips
(b)
F=
s
A
= 4 Nu 1
1
3
A bh 
= in

( )( )
57 .1 125
F 16 kips
==
4
==


4
2

2

 
.1
125

u
N
u
sA
(c)
F=
y
y
N
Ny = 5.2 11
3
A bh 
= in

==


4
2

2

 
.1 125

( )( )
42 .1 125
F 18 9. kips
==
2
10. A ¾-in.bolt, made of cold-finished B1113, has an effective stress area of 0.334 sq. in.
and an effective grip length of 5 in. The bolt is to be loaded by tightening until the
tensile stress is 80 % of the yield strength, as determined by measuring the total
elongation. What should be the total elongation?
Solution:
sL
δ=
E
from Table AT 7 for cold-finished B1113
s ksi y = 72
then, s s ( ) ksi = 80.0 y = 8.0 72 = 57 6.
E 30 10 psi 30,000 ksi 6
=⋅=
( )( )in
sL
.0 0096
57 6. 5
δ===
E
30,000
10
SECTION 1– DESIGN FOR SIMPLE STRESSES
11. A 4-lb. weight is attached by a 3/8-in. bolt to a rotating arm 14-in. from the center of
rotation. The axis of the bolts is normal to the plane in which the centrifugal force
acts and the bolt is in double shear. At what speed will the bolt shear in two if it is
made of AISI B1113, cold finish?
Solution:
From Table AT 7, s ksi psi
us = 62 = 62,000
1
= in
A2






()2


8
=π4
.0 2209
W
F = = us
g
2
4 ω=
32.2
( ) ( )( ) 14
2
3
2
ω
rsA
62,000 .0 2209
ω = 88 74. rad
sec 2
= = 60
88 74.
ω
πn
n = 847 rpm
12. How many ¾-in. holes could be punched in one stroke in annealed steel plate of AISI
C1040, 3/16-in. thick, by a force of 60 tons?
Solution:
For AISI C1040, from Figure AF 1
s ksi
u = 80
s s ( ) ksi ksi
us u = 75.0 = 75.0 80 = 60
A td 
= in
3

3
=π =π 4

16

  .0 4418 2

 
F = 60 tons =120
kips n = number of
holes
F
120
n holes = = =
4415 60
Asus
5
( )( ) .0
13. What is the length of a bearing for a 4-in. shaft if the load on the bearing is 6400 lb.
and the allowable bearing pressure is 200 psi of the projected area?
Solution:
pDL =W
where
p = 200 psi
D = 4 in
11
SECTION 1– DESIGN FOR SIMPLE STRESSES
W = 6400 lb
(200)(4)L = 6400
L = 8 in
BENDING STRESSES
DESIGN PROBLEMS
14. A lever keyed to a shaft is L =15 in long and has a rectangular cross section of h = 3t
. A 2000-lb load is gradually applied and reversed at the end as shown; the
material is AISI C1020, as rolled. Design for both ultimate and yield strengths. (a)
What should be the dimensions of a section at a =13 in ? (b) at b = 4 in ? (c) What
should be the size where the load is applied?
Problem 14.
Solution:
For AISI C1020, as rolled, Table AT 7
s ksi
u = 65
s ksi y = 49
Design factors for gradually applied and reversed load
= 8 Nu
Ny = 4
3
th
I = , moment of inertial
12
but h = 3t
36
h
4
I=
Moment Diagram (Load Upward)
12
SECTION 1– DESIGN FOR SIMPLE STRESSES
Based on ultimate strength
s
s=
u
u
N
(a) IFac
Mc
s==
I
h2
c=
F = 2000 lbs = 2 kips
s
65
36
h = 86.3 in
86.3
in h t 29.1 = = =
3
say
3
1
==


( )( ) 2 
13
h2
8

 
h
4

 
h in in
= 5.4 = 2
41
t in in
= 5.1 2
=1
(b) IFbc
Mc
s==
I
h2
c=
F = 2000 lbs = 2 kips
s
65
==


( )( ) 2 
4
8
h2

 
3
3
h
4

 
36
h = 61.2 in
in h
61.2
t 87.0 = = =
say
h=3
in t
=1in
(c)
13
SECTION 1– DESIGN FOR SIMPLE STRESSES
3
=
− h 4 5.4 3
h = 33.2 in
1
−t4
t = 78.0 in
say
h = .2 625 in or h in
−
=
5.1 1
−
13 4 −
=2
8
13 4 −
5
15. A simple beam 54 in. long with a load of 4 kips at the center is made of cast steel,
SAE 080. The cross section is rectangular (let h ≈ 3b ). (a) Determine the
dimensions for N = 3 based on the yield strength. (b) Compute the maximum
deflection for these dimensions. (c) What size may be used if the maximum
deflection is not to exceed 0.03 in.?
Solution:
For cast steel, SAE 080 (Table AT 6)
s ksi y = 40
E psi 6
= 30⋅10
14
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Table AT 2
Max. ( )( )kips in
FL
M = = = 54 − 4
54
bh
I=
4
4
3
12
but h = 3b
h
4
s
s
==
I=
36
(a) IMc
y
N
y
c=
h2


( ) 54 2
40
3
=
h



h
  4

 
3
5.1 1
36
h = 18.4 in
in h
18.4
b 39.1 = = =
3
3
= 4 , in in h
1
say h in
b
21
5.4
====
2
3
(b) ( )( )
33
FL
.0 0384
δ==
4000 54
48
()
( )( )in
=
3
48 30
EI
10 ⋅
3
6
12

  
5.1 5.4 
(c)
=
FL δ

h
  4
48
E

 
36
( )( ) ( )
3
4000 54 36 03.0
⋅h
48 30 10
=
( )( ) 6 4
h = 79.4 in
in h
79.4
b 60.1 = = =
3
3
= 25.5 = 5 , in in h
1
25.5
====
say h in in
b
43
4
3
3
75.1 1
15
SECTION 1– DESIGN FOR SIMPLE STRESSES
16. The same as 15, except that the beam is to have a circular cross section.
Solution:
s
(a) IMc
==
s
y
y
d
N4
π
I
d
2
d
=


sπ π
=
=
c=
64
2
M


43

d
 
d

  64
=
3
πd
()
40
32 54
d = 46.3
i
n
1
say d in
=3
2
3
FL
(b)
EI
δ=
π
I
=
48
d
4
64
( )( )
()
64
FL
δ
==
32
3
3
.0 0594
64 4000 54
48
( )( )( )
Ed
4
in
64
=
3
M
π π 48 30 10 5.3
⋅
FL
64
3
(c)
()4
π
δ=
48
Ed
( )( ) 3
64 4000 54 03.0
⋅πd
48 30 10
=
( )( ) 6 4
d = 15.4
i
n
1
say d in
=4
4
17. A simple beam, 48 in. long, with a static load of 6000 lb. at the center, is made of
C1020 structural steel. (a) Basing your calculations on the ultimate strength,
determine the dimensions of the rectangular cross section for h = 2b . (b)
Determine the dimensions based on yield strength. (c) Determine the dimensions
using the principle of “limit design.”
16
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
From Table AT 7 and Table 1.1
s ksi
u = 65
s ksi y = 48
= 3 ~ 4 Nu, say 4
Ny = 5.1 ~ 2 , say 2
( )( )in kips
FL
M = = = 72 −
6 48
4
Mc I
s=c=
h
2
4
3
bh
I=
12
but 2h b
=
h
h
4


s=
I=
24
2
=
h
M


43
12
M
h
24
(a) Based on ultimate strength
u
u
12
M3
s
s
==
=
4
h
65
N
()
12 72
h
3
h = 76.3 in
17
SECTION 1– DESIGN FOR SIMPLE STRESSES
76.3
in h b 88.1 = = =
2
2
= 75.3 = 3 , in in h
3
say h in in
b
87
75.3
====
4
2
2
(b) Based on yield
strength
.1 875 1
s
12
M
s
y
3
==
N
h
()
y
48
12 72
=
3
h
2
h = 30.3 in
in h
30.3
b 65.1 = = =
2
2
= 5.3 = 3 , in in h
1
say h in in
b
43
5.3
====
2
( )42
72 48
=
h = 29.2 in
in h
29.2
2
2
(c) Limit design (Eq.
1.6)
75.1 1
= 4
bh
Ms2 y
h

h


2

b .1 145 = = =
2
2
= 5.2 = 2 , in in h
1
5.2
====
say h in in
b
41
2
2
2
25.1 1
18. The bar shown is subjected to two vertical loads, F1 and F2, of 3000 lb. each, that are
L =10 in apart and 3 in. ( a , d ) from the ends of the bar. The design factor is 4
based on the ultimate strength; h = 3b . Determine the dimensions h and b if the
bar is made of (a) gray cast iron, SAE 111; (b) malleable cast iron, ASTM A4752, grade 35 018; (c) AISI C1040, as rolled (Fig. AF 1). Sketch the shear and
moment diagrams approximately to scale.
18
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problems18, 19.
Solution:
F F R R 3000 lb 1 = 2 = 1 = 2 =
Moment Diagram
M = R a = (3000)(3) = 9000 lbs −in = 9 kips −in 1
N = factor of safety = 4 based on u
s
1
bh
I=
2
3
h
2
c=

h
h


3



4
3
I=
=
h
12 36
(a) For gray cast iron, SAE 111
s ksi
u = 30 , Table AT 6
h
M

Mc

s
su =
===
43
N 30
18

h
36
  
h
 
18 9
I
()
4


2
s==
h
3
h = 78.2 in
in h
78.2
b 93.0 = = =
say h = 5.3 in , b =1in
3
3
(b) For malleable cast iron, ASTM A47-52, grade 35 018
M
19
SECTION 1– DESIGN FOR SIMPLE STRESSES
s ksi
u
= 55 , Table AT 6
M
h
Mc
s


2


()

h
36
  
h
 
18 9
18
M
s =
===
u
43
N 55
I
4
s==
h
3
3
1
3
3
h = 28.2 in
in h
28.2
b 76.0 = = =
= 2 , b in
(c) For AISI C1040, as rolled
s ksi
u = 90 , Fig. AF 1
h
M
Mc
s
su =
===
say h in 4
=
4


2


()

h
36
  
h
 
18 9
43
N 90
I
s==
18
M
4
h
h=
93.1 in 3
3
3
7
5
8
=
in h
93.1
b 64.0 = = =
say h in
=1 , b in
8
19. The same as 18, except that F1 acts up ( F2 acts down).
Solution:
[
]M
∑ =0
A
R R 1875 lb 1 = 2 =
20
SECTION 1– DESIGN FOR SIMPLE STRESSES
Shear Diagram
Moment Diagram
M =maximum moment = 5625 lb-in = 5.625 kips-in
(a) For gray cast iron
s
su = = M 3
18
=
N
18 .5
30
625
h
()
4
h
3
3
1
3
4
=
3
h = 38.2 in
in h
38.2
b 79.0 = = =
say h in
= 2 , b in
(b) For malleable cast iron
s
4
su = = M 3
18
=
N
18 .5
55
625
h
()
4
h
3
3
7
5
8
=
3
h = 95.1 in
in h
95.1
b 65.0 = = =
say h in
=1 , b in
8
21
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) For AISI C1040, as rolled
s
()
N
90
h
4
su = = M 3
18
=
18 .5
h
625
3
h = 65.1 in
in h
65.1
b 55.0 = = =
=1 , b in
3
31
1
say h =
in 2 2
20. The bar shown, supported at A and B , is subjected to a static load F of 2500 lb. at θ =
0 . Let d = 3 in , L =10 in and h = 3b . Determine the dimensions of the section if
the bar is made of (a) gray iron, SAE 110; (b) malleable cast iron, ASTM A47-52,
grade 32 510; (c) AISI C1035 steel, as rolled. (d) For economic reasons, the pins
at A, B, and C are to be the same size. What should be their diameter if the
material is AISI C1035, as rolled, and the mounting is such that each is in double
shear? Use the basic dimensions from (c) as needed. (e) What sectional
dimensions would be used for the C1035 steel if the principle of “limit design”
governs in (c)?
Problems 20, 21.
Solution:
22
SECTION 1– DESIGN FOR SIMPLE STRESSES
[
]M
∑ =0
A
3 =13(2500) RB
R lb B =10,833
[
]M
∑ =0
B
3 =10(2500) RA
R lb A = 8333
Shear Diagram
Moment Diagram
M = (2500 )(10) = 25,000 lb −in = 25 kips −in
h = 3b
bh
3
I=
12
4
Mc
s=
h
I = c = 36 h h
M
2


2
18

 M
==
43
I

h
  36

h
 
(a) For gray cast iron, SAE 110
s ksi
u = 20 , Table AT 6
N = 5 ~ 6 , say 6 for cast iron, dead load
18
M3
s
su = =
=
N
18 25
20
h
()
h
6
3
23
SECTION 1– DESIGN FOR SIMPLE STRESSES
h = 13.5 in
1
say h in
in h
b 71.1
3
==
3
= 5 , b in
4
4
=1
(b) For malleable cast iron, ASTM A47-32 grade 32510
s ksi
u = 52 , s ksi y = 34
N = 3 ~ 4 , say 4 for ductile, dead load
18
M3
s
su = =
=
N
18 25
52
h
()
4
h = 26.3 in 3
3
say h in
in h
b 09.1
==
h
3
1
= 3 , b in
=1
4
4
(c) For AISI C1035, as rolled
s ksi
u = 85 , s ksi y = 55
N = 4 , based on ultimate strength
18
M3
s
su = =
=
N
18 25
85
h
()
3
h
4
h = 77.2 in
in h
b 92.0
==
3
say h = 3 in , b =1in
(d) For AISI C1035, as rolled
s ksi
su = 64
N = 4 , R kips B =10.833
su B
s
s
s
R
N
A
==


 2 2
ππ
A2DD
=
42
=
64
10.833
==
s
657 in
s π 2
D = .0 D
4
2
24
SECTION 1– DESIGN FOR SIMPLE STRESSES
11
say D in
=
(e) Limit Design
bh
For AISI C1035 steel, s ksi y = 55

h3
b=

16
= y4
Ms2
h
M
==
( )43
8
=


h
2
25 55
h = 76.1 in
in h
b 59.0
7
==
5
3
say h in in
= .1 875 =1 , b in
8
21. The same as 20, except that o
θ = 30 . Pin B takes all the horizontal thrust.
Solution:
FV = F cosθ
[
]M
∑ =0
[
A
]M
∑ =0
B
RB FV
3 =13
3 =13(2500)cos30 RB
R lb B = 9382
RA FV
3 =10
3 =10(2500)cos30 RA
R lb A = 7217
Shear Diagram
25
SECTION 1– DESIGN FOR SIMPLE STRESSES
Moment Diagram
M = (2165 )(10) = 21,650 lb − in = 21 65. kips − in
18
M
s=
3
h
(a) For gray cast iron, SAE 110
s ksi
u = 20 , Table AT 6
N = 5 ~ 6 , say 6 for cast iron, dead load
18
M3
s
u
s ==
N
()
20
h
65.
=
18 21
6
h
3
h = 89.4 in say h in
in h
b 63.1
==
3
3
1
= 5 , b in
4
4
=1
(b) For malleable cast iron, ASTM A47-32 grade 32510
s ksi
u = 52 , s ksi y = 34
N = 3 ~ 4 , say 4 for ductile, dead load
18
M3
s
u
s ==
N
()
52
h
65.
=
18 21
3
h
4
h = 11.3 in
in h
b 04.1
==
3
say h = 3 in , b =1in
(c) For AISI C1035, as rolled
s ksi
u = 85 , s ksi y = 55
N = 4 , based on ultimate strength
26
SECTION 1– DESIGN FOR SIMPLE STRESSES
18
s
M3
su = =
N
85
h
()
4
h = 64.2 in
65.
=
18 21
3
h
in h
b 88.0
==
3
7
5
= 2 , b in
say h =
in 8 8
(d) For AISI C1035, as rolled
s ksi
su = 64
N = 4 , R lb BV = 9382
R F F lb BH H = = sinθ = 2500sin 30
=1250 ( ) ( )
2
2
=
+
=
+ 1250 RB RBV
9382
222
RBH R lb B = 9465
s
su B
s s= =

R
N
A

 2 2
ππ
A2DD
=
42
=
64
.9 465
s
say D
s π 2
D = .0 in
==
614 in D
4
2
5
8
=
(e) Limit Design
bh
For AISI C1035 steel, s ksi y = 55

h3
b=

= y4
Ms2
h


h
2
M
==
( )43
8
=
21 65. 55
h = 68.1 in
in h
b 56.0
7
==
5
3
say h in in
= .1 875 =1 , b in
8
27
SECTION 1– DESIGN FOR SIMPLE STRESSES
22. A cast-iron beam, ASTM 50, as shown, is 30 in. long and supports two gradually
applied, repeated loads (in phase), one of 2000 lb. at e =10 in from the free end,
and one of 1000 lb at the free end. (a) Determine the dimensions of the cross
section if b = c ≈ 3a . (b) The same as (a) except that the top of the tee is below.
Problem 22.
Solution:
For cast iron, ASTM 50
s ksi
u = 50 , s ksi
uc =164
For gradually applied, repeated load
N = 7 ~ 8 , say 8
M = F d + F (d + e) 1 2
where:
F 2000 lb 1 =
F 1000 lb 2 =
d = 30 −10 = 20 in
d + e = 30 in
M = (2000)(20)+ (1000)(30) = 70,000 lb −in = 70 kips − in
c
I
s=
M
Solving for I , moment of inertia




( )( ) ( )( ) [ ] ( )( ) ( )( ) a a a a y 3  = +

a
5
a
3
aa
2


aa33
2
2
+

3a
y=
28
SECTION 1– DESIGN FOR SIMPLE STRESSES
( )( )
( )( )
( )( )( )
( )( )( )2
3
a
3
4
3
aa
17
aa3
2
I=+++=
12
(a)
3
aaa
12
aaa
32
ct = cc
3a 2
5a 2
=
Based on tension
Mc
s
ut
s
t
==
N
I
3


50
8
=
a


( ) 70 2

17
 
a
4

 
2
a = .1 255 in
Based on compression
Mc
s
uc c
s
c
==
8


( ) 70 2
N


164
=

17
 
2
a = .1 001in
Therefore a = .1 255
in 1
Or say a in
And b = c = 3a = 3( 25.1 ) = 75.3 in
=1
I5
a
a
4

 
4
29
SECTION 1– DESIGN FOR SIMPLE STRESSES
3
Or b c in
==3 4
(b) If the top of the tee is below
=
c
5a
2
3a
2
17 4
a
2
t
=
c
c
=
I
M = 70 kips −in
Based on tension
s
Mc
su t t = =
8


( ) 70 2
N


50
=

17
 
I5
a
a
4

 
2
a = .1 488 in
Based on compression
Mc
s
uc c
s
c
==
8
2
a = .0 845 in
Therefore a = .1 488


( ) 70 2
N


164
=

17
 
I3
a
a
4

 
in 1
Or say a in
=1
2
And b c a in
1
==3 2
=4
CHECK PROBLEMS
30
SECTION 1– DESIGN FOR SIMPLE STRESSES
23. An I-beam is made of structural steel, AISI C1020, as rolled. It has a depth of 3 in.
and is subjected to two loads; F1 and 2 1 F = 2F ; F1 is 5 in. from one end and F2 is
5 in. from the other ends. The beam is 25 in. long; flange width is b = .2 509 in ; 4
I 9.2 in x = . Determine (a) the approximate values of the load to
cause elastic failure, (b) the safe loads for a factor of safety of 3 based on the yield
strength, (c) the safe load allowing for flange buckling (i1.24), (f) the maximum
deflection caused by the safe loads.
Problems 23 – 25.
Solution:
[
]M
∑ =0
( ) F F RB
A
5 20 2 25 1 + 1 =
1
[
RB = 8.1 F
]F
∑ =0
V
F1 + F1 = RA + RB
2
RA = 3F − 8.1 F = 2.1 F
Shear Diagram
111
Moment Diagram
31
SECTION 1– DESIGN FOR SIMPLE STRESSES
M = 9F = maximum moment
For AISI C1020, as rolled s
ksi y = 48
1
(a) IMc
s y=
where in d
3
c 5.1
===
( )( )
48 F1
2.9
s y= =
9 5.1
F 10 31. kips 1 =
F 2F 20 62. kips 2 = 1 =
s
(b)
IMc
sy
==
N
( )( )
48 F1
s = = 3 9 5.1
2
2
2.9
F 44.3 kips 1 =
F 2F 88.6 kips 2 = 1 =
L
(page 34)
25
(c) 96.9 15
2.509
b
==<
s ksi
c = 20 ( page 34, i1.24)
sc =
Mc
1
20 F =
9 5.1
2.9
F 30.4 kips 1 =
F 2F 60.8 kips 2 = 1 =
(d) For maximum deflection,
by method of superposition, Table AT 2
2
()
′
3
+′
=a L b
Fb

max
3 3 
y , a > b′


EIL
or
()
2
3

 ′ +
=b L a
Fa

max
3 3 
y , b′ > aEIL
I
( )( )
32
SECTION 1– DESIGN FOR SIMPLE STRESSES
=b L a
max y caused by F a
F1

1

()
2
3

 ′ +
y , 11
b′ > a
1111
max
33
EIL
where E = 30,000 ksi
a 5 in 1 =
b 20 in 1′ =
L = 25 in
F
()5

 + 5
( )( )( ) 3 30,000 9.2
25
Fb

2

max y caused by F2
2
3
a > b′
+′
=a L b
y , 22
2222


max
20 25
.0 0022 1F
1
′
()
2
=
()
4
3
y=

max
I = 9.2
in
33
EIL
3
1
where b 5 in 2′ =
a 20 in 2 =
()
25F
()

 +
20 25 5
3
y=

2
max .0 0043 2F
=
1
( )( )( )
1
3 30,000 9.2 25
3
Total deflection = δ
δ = y + y = F + F = F max max 1 1 0065 1
12
.0 022 .0 0043 .0
Deflection caused by the safe loads in (a)
( ) in a δ = .0 0065 10 31. = .0 067
Deflection caused by the safe loads in (b)
( ) in b δ = .0 0065 44.3 = .0 022
Deflection caused by the safe loads in (c)
( ) in c δ = .0 0065 30.4 = .0 028
24. The same as 23, except that the material is aluminum alloy, 2024-T4, heat treated.
Solution:
For aluminum alloy, 2024-T4, heat treated
s ksi y = 47
(a) IMc
s y=
33
SECTION 1– DESIGN FOR SIMPLE STRESSES
( )( )
47 F1
9 5.1
s y= =
2.9
F 10 10. kips 1 =
F 2F 20 20. kips 2 = 1 =
s
(b)
IMc
sy
==
N
( )( )
47 F1
2.9
s==3
9 5.1
F 36.3 kips 1 =
F 2F 72.6 kips 2 = 1 =
L
(page 34)
25
(c) 96.9 15
2.509
b
==<
s ksi
c = 20 ( page 34, i1.24)
sc =
Mc
I
( )( )
1
20 F =
9 5.1
2.9
F 30.4 kips 1 =
F 2F 60.8 kips 2 = 1 =
(d) Total deflection = δ
δ = y + y = F + F = F max max 1 1 0065 1
12
.0 022 .0 0043 .0
Deflection caused by the safe loads in (a)
( ) in a δ = .0 0065 10 10. = .0 066
Deflection caused by the safe loads in (b)
( ) in b δ = .0 0065 36.3 = .0 022
Deflection caused by the safe loads in (c)
( ) in c δ = .0 0065 30.4 = .0 028
25. A light I-beam is 80 in. long, simply supported, and carries a static load at the
midpoint. The cross section has a depth of d = 4 in , a flange width of b = 66.2 in ,
and 4
I 0.6 in x = (see figure). (a) What load will the beam support if it is made of
C1020, as-rolled steel, and flange buckling (i1.24) is considered? (b) Consider the
stress owing to the weight of the beam, which is 7.7 lb/ft, and decide whether or
not the safe load should be less.
34
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution:
(a) For C1020, as rolled, s ksi
u = 65
Consider flange buckling
80
==
Lb
2.66
30
L
since 15 < <
22 5.
40 b
22 5.
sc15
=
2 2=

ksi 30
=
()
1
 
L

1 1800 +
+
1800
b
s=
Mc I
in d
4
c2===
2
2
From Table AT 2
( )F
FL F
M 20
80
===
4
4
Mc
ss=c
= I
( )( )
15 F
=
20 2 6
F = 25.2 kips, safe load
(b) Considering stress owing to the weight of the beam
2
wL
M = (Table AT 2)
add’l
8
where w = 7.7 lb ft
35
SECTION 1– DESIGN FOR SIMPLE STRESSES
add’l ( )lb in kips in
wL
M  = − = − 22


7.7

80
8
M = 20F + .0 513 = total moment
12

8
ss=c
=
Mc I
= = 513 .0 513
( )( )
=F
15 +
20 .0 513 2
6
F = .2 224 kips
Therefore, the safe load should be less.
26. What is the stress in a band-saw blade due to being bent around a 13 ¾-in. pulley?
The blade thickness is 0.0265 in. (Additional stresses arise from the initial tension
and forces of sawing.)
Solution:
in t
c .0 0265 .0 01325
===
2
r =13 75. + .0 01325 =13.76325 in
Using Eq. (1.4) page 11 (Text)
s=
E
c
r
where E psi 6
= 30⋅10
( )( )
30 10 .0 01325 6=
⋅
s 28,881 psi
25
=
13.763
27. A cantilever beam of rectangular cross section is tapered so that the depth varies
uniformly from 4 in. at the fixed end to 1 in. at the free end. The width is 2 in. and
the length 30 in. What safe load, acting repeated with minor shock, may be
applied to the free end? The material is AISI C1020, as rolled.
Solution:
For AISI C1020, as rolled
s ksi
= 65 (Table AT 7)
Designing based on ultimate strength,
N = 6, for repeated, minor shock load
u
36
SECTION 1– DESIGN FOR SIMPLE STRESSES
65
s
s 10 8. = ksi
u
==
6
N
Loading Diagram
41−
−
30
h = 10.0 x
wh
+1 3
I=
=
h1x
12
c=
==
xFx
Mc
2
s
M = Fx( ) Fx 


 h
6
Fx
()
I
h2
3 3 Fx
===
3
222

wh 12 
2
 
  h
h
10.0 1 +
Differentiating with respect to x then equate to zero to solve for x giving maximum stress.
( ) ( ) ( )( )( )
ds
 
 +

2
10.0 1 1 2 10.0 1 10.0
xxx
+−+
=
x
dx
34
F
0
()
h = 10.0 (10)+1
=
10.0 x +1− 2( = 2 in
10.0 x) = 0 x 10.0 1
=10 in

3
s
su = =
N
h
3 10
10 8.F
Fx 2
()
=
( )2 2
F = 44.1 kips
TORSIONAL STRESSES
DESIGN PROBLEMS
37
SECTION 1– DESIGN FOR SIMPLE STRESSES
28. A centrifugal pump is to be driven by a 15-hp electric motor at 1750 rpm. What
should be the diameter of the pump shaft if it is made of AISI C1045 as rolled?
Consider the load as gradually repeated.
Solution:
For C1045 as rolled,
s ksi y = 59
s ksi
us = 72
Designing based on ultimate strength
s
us
s = , N = 6 (Table 1.1)
N
72
s 12 ksi
==
6
Torque, ( )
hp
T = = = 45 − = 540 − = .0 540 −
33,000
33,000 15
2 1750
2
ft lb in lb in kips
()
n
ππ
For diameter,
16
=
d
sπ
16 .0 540
12π d
d = .0 612
in
5
say d in
T
()
3
=
3
=
8
29. A shaft in torsion only is to transmit 2500 hp at 570 rpm with medium shocks. Its
material is AISI 1137 steel, annealed. (a) What should be the diameter of a solid
shaft? (b) If the shaft is hollow, Do Di = 2 , what size is required? (c) What is the
weight per foot of length of each of these shafts? Which is the lighter? By what
percentage? (d) Which shaft is the more rigid? Compute the torsional deflection
of each for a length of 10 ft.
Solution:
hp
()
T = = = 23,036 − = 276 −
33,000
2500
33,000
2
ft lb in kips
()
n
2 570
ππ
For AISI 1137, annealed
s ksi y = 50 (Table AT 8)
s s ksi ys = 6.0 y = 30
Designing based on yield strength N =
3 for medium shock, one direction
38
SECTION 1– DESIGN FOR SIMPLE STRESSES
Design stress
s
30
N
3
sys 10 = = ksi
=
(a) Let D = shaft diameter
4
D
=
s=
T
c
J
π
J
c=
32
D
2
T
=
16
D
3
sπ
()
16 276
10π D
=
3
=5
4
2
−
D = 20.5
i
n
1
say D in
(b) (
) [( ) ]32 4 4 4 4 4 Do Di Di Di Di
−
πππ
J
15
=
32
=
32
=
2
TD
2
32
T
DD
2
c===
oi
D
i
sπ π
=
=
i
43


15
   
D
15 i i
D
32
()
32 276
10π Di
=
3
15
D in i = 66.2
D D in o i = 2 = 32.5
say
i
D in 85
=2
o
D in 41
=5
(c) Density, 3
ρ = .0 284 lb in (Table AT 7)
39
SECTION 1– DESIGN FOR SIMPLE STRESSES
For solid shaft
w = weight per foot of length


w D 3 D 3 ( )( ) .0 284 25.5 73 8. lb ft 12
= πρ π
ρ
===4

π
22
2

For hollow shaft
π
w(
) ( ) D D D D ( ) ( ) ( ) [ ] lb ft o i 3 o i 3 .0 284 25.5 .2 625 55 3.


12
2222
ρ
22
= πρ π

−=−=−=4

Therefore hollow shaft is lighter
−
73 8. 55 3.
=
Percentage lightness = ( ) 100% 33 5. % 55.3
(d) Torsional Deflection
TL
θ=
JG
where
L =10 ft =120 in
G ksi 3
=11 5. ⋅10
4
π
J
For solid shaft,
32
=
180
( )( )
D

.0 039 .0 039 

276 120
()o
()()
θ rad
=π π



4
3

2.2
==
=
⋅
32

For hollow shaft, (
25.5 11
5. 10
) 4 4 Do Di
J−
π
=
( )( ) 276 120
180

.0 041 .0 041 
()o
[]()()()
θ rad
=π π
32




44

4.2
==
=
3
−⋅
32
5. 10 
25.5 .2 625 11
Therefore, solid shaft is more rigid, o o
2.2 < 4.2
30. The same as 29, except that the material is AISI 4340, OQT 1200 F.
Solution:
hp
()
T = = = 23,036 − = 276 −
33,000
2500
33,000
2
ft lb in kips
()
n
2 570
ππ
For AISI 4340, OQT 1200 F
s ksi y =130
s s ( ) ksi ys = 6.0 y = 6.0 130 = 78
Designing based on yield strength
40
SECTION 1– DESIGN FOR SIMPLE STRESSES
N = 3 for mild shock
Design stress
s
78
N
3
s 26 = = ksi
ys
=
(a) Let D = shaft diameter
s=
T
c
J
π
J
4
D
=
c=
32
D
=
16
T
16 276
26π D
2
()
sπ
D
3
=
3
=3
4
2
−
D = 78.3
i
n
3
say D in
(b) (
) [( ) ]32 4 4 4 4 4 Do Di Di Di Di
−
πππ
J
=
32
=
2
TD
2
32
T
15
=
32
DD
2
c===
oi
D
i
=
=
sπ π
i
43


15
   
D
15 i i
D
32
()
32 276
26π Di
D in i = 93.1
D D in o i = 2 = 86.3
say
D in i = 2
D in o = 4
=
15
3
(c) Density, 3
ρ = .0 284 lb in (Table AT 7)
41
SECTION 1– DESIGN FOR SIMPLE STRESSES
For solid shaft
w = weight per foot of length


w D 3 D 3 ( )( ) .0 284 75.3 37 6. lb ft 12
= πρ π
ρ
===4

π
22
2

For hollow shaft
π
w(
) ( ) D D D D ( ) ( ) ( ) [ ] lb ft o i 3 o i 3 .0 284 4 2 32 1.


12
2222
22

= πρ π
ρ

−=−=−=4
Therefore hollow shaft is lighter
37 6. 32 1.
=
−
Percentage lightness = ( ) 100% 17 1. % 32.1
(d) Torsional Deflection
TL θ =
JG
where
L =10 ft =120 in
G ksi 3
=11 5. ⋅10
π
J
For solid shaft,
32
4
D
=

( )( )
276 120
180

.0 148 .0 148 
()
()() o
θ rad
=π π



4
3

48.8
==
=
⋅
32

For hollow shaft, (
) 4 4 Do Di
J−
π
=
75.3 11
5. 10
( )( ) 276 120



32
180
.0 122 .0 122
()o
[]()()()
θ rad
=π π



44
3
−⋅
32
4 2 11
5. 10

==
=

99.6
Therefore, hollow shaft is more rigid, o o
99.6 < 48.8 .
31. A steel shaft is transmitting 40 hp at 500 rpm with minor shock. (a) What should be
its diameter if the deflection is not to exceed 1o in 20D ? (b) If deflection is
primary what kind of steel would be satisfactory?
Solution:
(a) ( )
hp
T = = = 420 − = 04.5 −
33,000
33,000 40
2
ft lb in kips
()
n
2 500
ππ
3
G ksi
=11 5. ⋅10
L = 20D
42
SECTION 1– DESIGN FOR SIMPLE STRESSES
π
1 o
θ = = JG
180
TL θ = rad
=
DD 04.5
π
20
( )( )
180⋅ 
 

π
  32
4
( ) 3 11
D = 72.1
i
n
3
say D in
=1
(b) ( )
16 16 04.5
T
s 8.4 = = =
()
ksi
4
5. 10
D
75.1
33
ππ
Based on yield strength
N=3
s Ns ( )( ) ksi ys = = 3 8.4 =14 4.
s
sys
y
14 4.
24 = = = ksi
0.6
Use C1117 normalized steel s ksi y = 35
0.6
32. A square shaft of cold-finish AISI 1118 transmits a torsional moment of 1200 in lb.
For medium shock, what should be its size?
Solution:
For AISI 1118 cold-finish
s ksi y = 75
s s ksi ys = 6.0 y = 45
N = 3 for medium shock
s
T
N
Z
9
9
2
b
==
4
sys′ = =
where, h = b
22 3
bhb
2
Z′ = = (Table AT 1)
T =1200 in − lb = 2.1 in − kips
()
45
s = = 2.1 9
3
b = h = 71.0
in
3
say b h in
3
43
SECTION 1– DESIGN FOR SIMPLE STRESSES
CHECK PROBLEMS
33. A punch press is designed to exert a force sufficient to shear a 15/16-in. hole in a
½-in. steel plate, AISI C1020, as rolled. This force is exerted on the shaft at a
radius of ¾-in. (a) Compute the torsional stress in the 3.5-in. shaft (bending
neglected). (b) What will be the corresponding design factor if the shaft is made
of cold-rolled AISI 1035 steel (Table AT 10)? Considering the shock loading that
is characteristics of this machine, do you thick the design is safe enough?
Solution:
For AISI C1020, as rolled
s ksi
us = 49
F s ( Dt) = us π
15
where D in
=
16 1
=
2
t in
15
49  =


F ( ) 72 2. kips
=π
where r in
T()
= in − kips

 
 


1
T = Fr
=
4
16 3
3
= 54 2.
72 2. 4
16
T
(a) 3
=
sπ
where d = 5.3 in
d

2

()
()
16 54 2.
s 44.6 ksi
==
π
3
5.3
(b) For AISI 1035 steel, s ksi
us= 64
for shock loading, traditional factor of safety, N =10 ~ 15
64
s
Design factor , 94.9
Nus , the design is safe ( N ≈10) = = =
s
6.44
34. The same as 33, except that the shaft diameter is 2 ¾ in.
Solution:
44
SECTION 1– DESIGN FOR SIMPLE STRESSES
d = 75.2 in
16
T
(a) 3 sπ
=
d
()
16 54 2.
s 13 3. ksi
()
==
π
75.2
3
(b) For AISI 1035 steel, s ksi
us= 64
for shock loading, traditional factor of safety, N =10 ~ 15
64
s
Design factor , 8.4
Nus , the design is not safe ( N <10 ) = = =
s
13.3
35. A hollow annealed Monel propeller shaft has an external diameter of 13 ½ in. and an
internal diameter of 6 ½ in.; it transmits 10,000 hp at 200 rpm. (a) Compute the
torsional stress in the shaft (stress from bending and propeller thrust are not
considered). (b) Compute the factor of safety. Does it look risky?
Solution:
For Monel shaft,
us= 98 (Table AT 3)
s ksi
N = 3 ~ 4 , for dead load, based on ultimate strength
(a) JTc
s=
( ) [( ) ( ) ] 4
4444
32in D D
−
ππ
=
Jo i =
=
13 5. 5.6 −
13 5.
co75.6 = = = 32
3086
in D
()
2
2
hp
T = = = 262,606 − = 3152 −
33,000
10,000
33,000
2
ft lb in kips
()
n
2 200
ππ
( )( )
3152 75.6
s 9.6 ksi
==
3086
(b) Factor of safety,
98
s
us
N , not risky
===
14 2.
6.9
s
45
SECTION 1– DESIGN FOR SIMPLE STRESSES
STRESS ANALYSIS
DESIGN PROBLEMS
36. A hook is attached to a plate as shown and supports a static load of 12,000 lb. The
material is to be AISI C1020, as rolled. (a) Set up strength equations for
dimensions d , D , h , and t . Assume that the bending in the plate is negligible. (b)
Determine the minimum permissible value of these dimensions. In estimating the
strength of the nut, let D 2.1 d 1 = . (c) Choose standard fractional dimensions
which you think would be satisfactory.
Problems 36 – 38.
Solution:
s = axial stress
s = shear stress
s
(a)
F
4
F
sπ
==
1
d
2
π
4
d
2
Equation (1) sF dπ4
=
46
SECTION 1– DESIGN FOR SIMPLE STRESSES
44F
F
F
4
F
s
− =π π π
=
=
()()[]()()
1
2
2
=
222
Dd
2
2
−
2.1
DD
2
44.1 D D
Dd−
4
π
−
1
4
d
1
Equation (2) 2
D = + 44.1
F
dh
s
s π 1 2.1 π
F
==
Dh
Equation (3)
h
dss 2.1 π
=
F
π
s
F
F
s
sπ
=
Dt
Equation (4)
F
Dss
tπ
=
(b) Designing based on ultimate strength,
Table AT 7, AISI C1020, as rolled u = 65
s ksi
s ksi
us = 49
N = 3 ~ 4 say 4, design factor for static load
65
s
su16 = = ksi
=
sus
4
N
s
12 = = =
49
ksi
s
N
F =12,000 lb =12 kips
From Equation (1)
()
4 4 12
F
d 98.0 = = =
ππs
()
in
16
From Equation (2)
()
4 22
F
4 12
44.1
()
d ( ) in
4
D 44.1 98.0 53.1 = + = + =
Equation (3)
16
s
ππ
From
F
12
h
= = = ( )( )
in
2.1
2.1
ds
ππ
98.0 12
27.0
s
From Equation (4)
= = = ( )( )
Ds
π πs
12
F
t
in
53.1 12
21.0
47
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) Standard fractional dimensions
d =1in
1
D in
=1 1
2
=
4
1
=
4
h in
t in
37. The same as 36, except that a shock load of 4000 lb. is repeatedly applied.
Solution:
(a) Same as 36.
(b) N =10 ~ 15 for shock load, based on ultimate strength
say N =15, others the same.
65
s
su4 = = =ksi
N
s
sus
15
49
N
15
3 = = = ksi
s
F = 4000 lb = 4 kips
From Equation (1)
()
444
F
d 13.1 = = =
ππs
()
in
4
From Equation (2)
()
4 22
F
()
44
d ( ) in
D 44.1 13.1 76.1 = + = + =
s
ππ
From Equation (3)
h
= = = ( )( )
44.1
4
F
4
in
13.1 3
31.0
ds
2.1
2.1
ππ
s
From Equation (4)
t
F
4
= = = ( )( )
Ds
π πs
in
76.1 3
24.0
48
SECTION 1– DESIGN FOR SIMPLE STRESSES
(c) Standard fractional dimensions
1
d in
=1
8
3
=1 3
4
=
81
D in
h in
=
t in
4
38. The connection between the plate and hook, as shown, is to support a load F .
Determine the value of dimensions D , h , and t in terms of d if the connection is
to be as strong as the rod of diameter d . Assume that D 2.1 d 1 = , us u
s = 75.0 s ,
and that bending in the plate is negligible.
Solution:
s
=
F
1
d
4
π
2
1
F d s2
=π
4
=
Ns
1π
 
(1)  
F d2 u
4
49
SECTION 1– DESIGN FOR SIMPLE STRESSES
=
=
F
F
s
1
Dd
1
( ) ( ) 2 2 2 244.1
DD
−
ππ
−
4
F (D d )s 2 2
1
1
4
= π − 444.1
=−
Ns
(2) (
1π
 
 
)
22u
FDd
4
F
44.1
==
Dh
sF =
2.1 π
dhs
F
s
s π1
dh
2.1 π
=
Ns
s
F dh
us
75.0 u


2.1 π 2.1 π

=


dh
N



=
Ns
F dh 5 u

(3) 
9.0 π 

F
s
sπ
=
Dt

=

Dt
F = π Dtss
=
us
u
F Dt 75.0 π π Ns


=
Ns

(4) 


F Dt u
75.0 π
s
N




Equate (2) and (1)
=−
Ns
s
1π π
()
1




22u2u
FDd
4
22
Equate (3) and (1) 1
=
Ns
F dh u 2 u 9.0 π π
D=



44.2 d 44.1
D = .1 N
562d  =

4
d


s



d
N
=

4
d


h .0 278
==
()
d
4 9.0
Equate (4) and (1)
=
Ns




1
s

F Dt u 2 u
d
=
75.0 π π

N


4
=
Ns

( )( ) 
1
s





F d tu 2 u
=
d
75.0 π .1 562 π

N
d
4
t .0 214 = =
d
( )( ) 4 75.0 .1 562
50
SECTION 1– DESIGN FOR SIMPLE STRESSES
39. (a) For the connection shown, set up strength equations representing the various
methods by which it might fail. Neglect bending effects. (b) Design this
connection for a load of 2500 lb. Both plates and rivets are of AISI C1020, as
rolled. The load is repeated and reversed with mild shock. Make the connection
equally strong on the basis of yield strengths in tension, shear, and compression.
Problems 39, 40
Solution:
(a)


=
2
ss
5D
4
π
D
5π4
Equatio F
n (1) s
−2
=
Fs
s
=
t( ) b D F
Equation (2) D
b=+2
F


1
ts
F
=
s
5
Dt
F
Equation (3) Ds
t
5
=
(b) For AISI C1020, as rolled
s ksi y = 48 (Table AT 7)
s s ksi ys = 6.0 y = 28
N = 4 for repeated and reversed load (mild shock) based on yield strength
48
s 12 ksi
==
4
28
s ksi
s7
==
4
From Equation (1)
51
SECTION 1– DESIGN FOR SIMPLE STRESSES
D
5π4
=
Fs
where
F = 2500 lb = 5.2 kips
( ) 4 5.2
4
F
s
5
π π say in
D
===
()
in
5
From Equation (3)
s
s
57
30.0
16
5.2
F
= = say in t 13.0
5
in
5=
Ds
5


5
16


( ) 12 32
F
5.2


5

From Equation (2)
= + = say 2 in b 96.1 2  =
D in
ts


5
32


( ) 12 +
2
16

40. The same as 39, except that the material is 2024-T4, aluminum alloy.
Solution:
(a) Same as 39.
(b) ) For 2024-T4, aluminum alloy
s ksi y = 47 (Table AT 3)
s s ksi ys = 55.0 y = 25
N = 4 for repeated and reversed load (mild shock) based on yield strength
47
s 12 ksi
==
4
25
==
s ksi
s6
4
From Equation (1)
D
5π4
Fs
=
where s
F = 2500 lb = 5.2 kips
( ) 4 5.2
4
F
π π say in
3
D
===
()
in
5
56
s
33.0
s
8
From Equation (3)
52
SECTION 1– DESIGN FOR SIMPLE STRESSES
5.2
1
F
= = say in t 11.0
in
5=


Ds
5
From Equation (2)
F
5.2
= + = say in b 42.2
2=
D in
ts


18


38




+
( ) 12 2
( ) 12 8

3
8
1

2
2
41. (a) For the connection shown, set up strength equations representing the various
methods by which it might fail. (b) Design this connection for a load of 8000 lb.
Use AISI C1015, as rolled, for the rivets, and AISI C1020, as rolled, for the
plates. Let the load be repeatedly applied with minor shock in one direction and
make the connection equally strong on the basis of ultimate strengths in tension,
shear, and compression.
Problem 41.
Solution:
(a)
P
s −
34
F
F
P
= or t( ) b D s 2
= Equation (1) −
t( ) b D
F
1
Equation (2)
ssR
=

42 

4

D
()2
π
53
SECTION 1– DESIGN FOR SIMPLE STRESSES
F
Dt
= Equation (3)
R
s 4
(b) For AISI C1015, as rolled
s ksi uR = 61 , s s ksi usR uR = 75.0 = 45
For AISI C1020, as rolled
s ksi uP = 65
N = 6, based on ultimate strength
s
suP
65
10 8. = = ksi
P
=
N
s
suR
6
61
N
s
susR
6
45
N
6
2 5.7
.0 412
7
16


10 1. = = ksi
R
=
sR
5.7 = = =ksi
F = 8000 lb = 8
kips Solving for D
F
sR
s π
=
2
2DF
7
8
π π say in
D
===
()
in
s
2
Solving for t
F
R
s 4
=
F
Dt
sR
16
1
8
= = say in t
4=
.0 453
Ds
R
Solving for b F
P
s −
=
Using
t( ) b D
in
4


( ) 10 1.
2
F
8
+=7
= + = say 2 in b
D in


ts
P
Using
12


( ) 10 8.
34
F
=
−
16
92.1
P
t( ) b D s 2
54
SECTION 1– DESIGN FOR SIMPLE STRESSES
()
3
=
= + = say 2 in b
F
38


7

2
D in
4
ts
P
1
4
2




( ) 10
8.
+
2
16

99.1
Therefore
b = 2 in
7
D in
=
16
=
2
1
t in
42. Give the strength equations for the connection shown, including that for the shear of
the plate by the cotter.
Problems 42 – 44.
Solution:
Axial Stresses
4
F
F
2
s
π
= = Equation (1)
1
π
D
D
2
1
4
F
1
s
−2
= Equation (2) ( ) L D e
55
SECTION 1– DESIGN FOR SIMPLE STRESSES
F
s
= Equation (3) D e
2
4
F
F
s =
− π
=
Equation (4)
()()
π
aD
2
−
s
1
2
aD
2
2
2
F
2
4
4
F
=
π
=
1
−
π
2
DDe
2
DDe
Equation (5)
4
−
2
2
2
2
4
Shear Stresses
F
eb
= Equation (6)
s
s2
F
s
s −+
=
2 2 Equation (7) ( ) L D e t
56
SECTION 1– DESIGN FOR SIMPLE STRESSES
F
at
s
sπ
= Equation (8)
F
Dm
= Equation (9)
s
s π1
F
Dh
s
s 22
= Equation (10)
43. A steel rod, as-rolled AISI C1035, is fastened to a 7/8-in., as-rolled C1020 plate by
means of a cotter that is made of as-rolled C1020, in the manner shown. (a)
Determine all dimensions of this joint if it is to withstand a reversed shock load F
=10 kips , basing the design on yield strengths. (b) If all fits are free-running fits,
decide upon tolerances and allowances.
Solution: (See figure of Prob. 42)
7
= = , sy y
t in .0 875 in 8 s = 6.0 s
For steel rod, AISI C1035, as rolled
s ksi y 55
1=
s ksi sy 33
1=
For plate and cotter, AISI C1020, as rolled
s ksi y 48
2=
s ksi sy 28
2=
N = 5 ~ 7 based on yield strength
say N = 7
From Equation (1) (Prob. 42)
()
N
D
s
41
F
syπ
==
4 10
7
πD
2
55
=
2
1
1
D 27.1 in 1
1
= 1
1=
say D in 4
57
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Equation (9)
s
ssy
==
s
1
F
33
N
7
π
10
Dm1
=
π


m = 54.0
in
9
say m in
1
14

=
m

16
From Equation (3)
.1 273 D2e =
From Equation (5)
F
s
sy
De2
1
==N
==7 De2
s
10
55
sy
41
s
==
πN
2
F
4
DDe
2
D 80.1 in 2 =
3
1 2=
say D in 4
and 273 .1 D2e
3
= 1=

e
e = 73.0
in
3
say e in
By further adjustment
5
Say D 2 in 2 = , e in
=
10
πD ()4
55
2
−
273
2−
7
2
4( ) .1
.1 273  
4
=
4
=
(8)
From
8
Equation
s
ssy
==2
28 7
=
πa
F
s
πN
at
10
( ) .0
875
a = 91.0 in
say a =1in
58
SECTION 1– DESIGN FOR SIMPLE STRESSES
From Equation (4)
sy−
==
()
πN
4 10
( ) 22 =π a
2
42
s
F
( ) 22
aD
48 7
−
a=
42.2 in
2
=2
21
=2
2
1
say a in
use a in
From Equation (7)
s
F
sy
s
==
22
2
−+N
()LDet
s
28
=
58
+
7
  22L 
 −

L = 80.2 in
10
( ) .0
875
say L = 3 in
From Equation (6)
7
b = 2 in
From Equation (10)
s
s
2N
sy
s
2
= = eb
F
10
=
28

2
58

s
ssy
b
 
==
F
2
s
22
28
2( ) 2 D h 5
=7
h
N
10
h in in
= .0 8
625 =
Summary of Dimensions
L = 3 in
5
h in
=
8
=
8
b=2
in 7
t in
59
SECTION 1– DESIGN FOR SIMPLE STRESSES
9
=
m in 16
1
a in
=2
2
=
8
1
1 1=
D in 4
D 2 in 2
=5
e in
(b) Tolerances and allowances, No fit, tolerance = ± .0 010
in L = 3 ± .0 010 in
h = .0 625 ± .0 010 in
t = .0 875 ± .0 010 in
m = .0 5625 ± .0 010 in
a = .2 500 ± .0 010 in
D 25.1 .0 010 in 1 = ±
For Free Running Fits (RC 7) Table 3.1
Female Male
0.2
.0 0030 0.2
.0 0040
−−
−+
b in
= b in
=
.0 0058
.0 0000
allowance = 0.0040 in
.0 0030
.0 0040
0.2
0.2
2
2
−−
−+
= D in
=
allowance = 0.0040 in
D in
.0 0058
.0 0000
.0 0016 .0 0020
.0
625
.0 625
+
−−
−
e in
= e in
=
.0 0030
.0 0000
allowance = 0.0020 in
44. A 1-in. ( D1) steel rod (as-rolled AISI C1035) is to be anchored to a 1-in. steel plate
(as-rolled C1020) by means of a cotter (as rolled C1035) as shown. (a) Determine
all the dimensions for this connection so that all parts have the same ultimate
strength as the rod. The load F reverses direction. (b) Decide upon tolerances and
allowances for loose-running fits.
Solution: (Refer to Prob. 42)
(a) For AISI C1035, as rolled
s ksi u 85
1=
s ksi us 64
1=
For AISI C1020, as rolled
60
SECTION 1– DESIGN FOR SIMPLE STRESSES
s ksi u 65
2=
s ksi us 48
2=
Ultimate strength
Use Equation (1)
1 22



F s D ( ) ( ) kips u u 1 66 8. = π π
11
=
4
on (9)
=π
Fusus
Equati D m
1
1
66 8. = (64)(π
)(1)m m = 33.0 in
3
say m in


1
=

85
4

=
8
1


From Equation (3)
F s D e uu21
=
( )D e2
66 8. = 85
.0 7859 D2e =
From Equation (5)
π
F = s D − D e uu2


2
1
4


( )  
66 8. 85 2
π D2
= − .0 7859 4
1
2

D 42.1 in 2 =
3
1 2=
say D in
8


D e = e 2 1 .0 7859

3
=
e = 57.0
8

=
16
in
9
say e in
From Equation (4)


1
=−

(
)

2
2
Fusu π a D
2
  
()
    


2


2
4
1
66 8. 65 π a
2
=
a=
say a in


3
1
79.1 in  − 
8
3
4
=1
4
From Equation (8)
61
SECTION 1– DESIGN FOR SIMPLE STRESSES
F s at u us π2
=
66 8. = (48)(π
)(a)(1) a = 44.0 in
1
say a in
=
2
3
use a in
=1
4
From Equation (2)
F s (L D )e u u 2 2
=−
=−
169
3

( ) 

20.3 in  
 1

66 8.
say L 
65 L 1
in
8
L=
4
=3
From
Equation
(7) F s (L 3
D e)t
9
u = us − 2 − 2
2
 
66 8. 2 48 1  
()()1
=L−−
8
say L in
L = 51.1 in 16
1
21
=1
use L in
=3
4
From Equation (6)
Fusus eb
=2
=
169
b
( )  


66 8. 2 64
b = 93.0 in
1
say b =1in
From Equation (10)
Fusus D2h
=2
1
=
83
h
( )  


66 8. 2 64 1
h = 38.0 in
3
say h in
=
8
=3
4
Dimensions
1
L in
62
SECTION 1– DESIGN FOR SIMPLE STRESSES
3
=
h in 8
b =1in
t =1in
3
m in
=
8
3
=1
4
=
16
a in
D 1in 1
3
= 1 2=
D in 8
9
e in
(b) Tolerances and allowances, No fit, tolerance = ± .0 010
in L = 25.3 ± .0 010 in
h = .0 375 ± .0 010 in
t = .1 000 ± .0 010 in
m = .0 375 ± .0 010 in
a = 75.1 ± .0 010 in
D .1 000 .0 010 in 1 = ±
For Loose Running Fits (RC 8) Table 3.1
Female Male
0.1
0.1
.0 0035
.0 0045
−−
−+
b in
= b in
=
.0 0065
.0 0000
allowance = 0.0045 in
.0 0040
.0 0050
.1
375
.1 375
2
2
−−
−+
D in
= D in
=
.0 0075
.0 0000
allowance = 0.0050 in
.0 0028
.0 0035
.0
5625
.0 5625
−−
−+
e in
= e in
=
.0 0051
.0 0000
allowance = 0.0035 in
45. Give all the simple strength equations for the connection shown. (b) Determine the
ratio of the dimensions a , b , c , d , m , and n to the dimension D so that the
connection will be equally strong in tension, shear, and compression. Base the
calculations on ultimate strengths and assume us u
s = 75.0 s .
63
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problems 45 – 47.
Solution:
(a) Neglecting bending

Equation (1): 
1


=2
FsπD
4
Equation (2): 

 
 
=2
F s 2 c sπ
4
Equation (3): F = s(2bc)
Equation (4): F = s(ac)
Equation (5): F = s[2 (d −
c)b] Equation (6): F s ( mb) s
= 4 Equation (7): F s ( nb) s =
2 Equation (8): F = s(d − c)a
(b) Ns
su = and Ns
sus
s=
Therefore
s s s = 75.0
Equate (2) and (1)


1
  = 


=2 2 F ss 2 π c s π D



1
4


1
1
4





2
=

4
1

 2 2


75.0 s c s D
c = .0 8165D
Equate (3) and (1)
1

( ) 


2
==
F s 2bc s π D 4
1
()2
2b .0 8165D = π D
4
b = .0 4810D
64
SECTION 1– DESIGN FOR SIMPLE STRESSES
Equate (4) and (1)


= =2
F sac s π D 4
1
()
2
a .0 8165D = π D
4
a = .0 9619D
Equate (5) and (1) 1

[ ] ( ) 


1


= − =2
Fs2dcbsπD4
1
( )( )
2
2 d − .0 8165D .0 4810 = π D
4
d = .1 6329D
Equate (6) and (1)
1

( ) 


2
==
F ss 4mb s π D
4
1
( )( ) 2
75.0 4m .0 4810D = π D
4
m = .0 5443D
Equate (7) and (1)
1

( ) 


2
==
F ss 2nb s π D
4
1
( )( ) 2
75.0 2n .0 4810D = π D
4
n = .1 0886D
Equate (8) and (1)
1

( ) 


2
=−=
FsdcasπD 4
1
()
2
.1 6329 − D − .0 8165D a = π
D4
a = .0 9620D
Summary
a = .0 9620D
b = .0 4810D
c = .0 8165D
d = .1 6329D
m = .0 5443D
n = .1 0886D
65
SECTION 1– DESIGN FOR SIMPLE STRESSES
46. The same as 45, except that the calculations are to be based on yield strengths. Let s =
6.0 s .
sy y
Solution: (Refer to Prob. 45)
(a) Neglecting bending

Equation (1): 
1


=2
FsπD
4
Equation (2): 

 
 
=2
F s 2 c sπ
4
Equation (3): F = s(2bc)
Equation (4): F = s(ac)
Equation (5): F = s[2 (d −
c)b] Equation (6): F s ( mb) s
= 4 Equation (7): F s ( nb) s =
2 Equation (8): F = s(d − c)a


1
(b)
Ns
= and
y
s
Ns
ssy s =
Therefore
s s s = 6.0
Equate (2) and (1)
=2 2 F ss 2 π c s π D
  = 





1
4


1
1
4





2
=

4


1
1

 2 2


6.0 s c s D
c = .0 9129D
Equate (3) and (1)
1

( ) 


2
==
F s 2bc s π D 4
1
()2
2b .0 9129D = π D
4
b = .0 4302D
Equate (4) and (1)
= =2
F sac s π D 4
1
()
2
a .0 9129D = π D


a = .0 8603D
4
66
SECTION 1– DESIGN FOR SIMPLE STRESSES
Equate (5) and (1) 1

[ ] ( ) 


2
=−=
Fs2dcbsπD
4
( )( )
1
2
2 d − .0 9129D .0 4302 = π D
4
d = .1 8257D
Equate (6) and (1)
1

( ) 


2
==
F ss 4mb s π D
4
1
( )( ) 2
6.0 4m .0 4302D = π D
4
m = .0 7607D
Equate (7) and (1)
1

( ) 


2
==
F ss 2nb s π D
4
1
( )( ) 2
6.0 2n .0 4302D = π D
4
n = .1 5214D
Equate (8) and (1)
1

( ) 
= − =2
FsdcasπD


4
()
1
2
.1 8257 − D − .0 9129D a = π D
4
a = .0 8604D
Summary
a = .0 8604D
b = .0 4302D
c = .0 9129D
d = .1 8257D
m = .0 7607D
n = .1 5214D
47. Design a connection similar to the one shown for a gradually applied and reversed
load of 12 kips. Base design stresses on yield strengths and let the material be
AISI C1040 steel, annealed. Examine the computed dimensions for proportion,
making changes that you deem advisable.
Solution: (See figure in Prob. 45 and refer to Prob. 46)
67
SECTION 1– DESIGN FOR SIMPLE STRESSES
N = 4 based on yield strength for gradually applied and reversed load.
For AISI C1040, annealed
s ksi y = 47 (Fig. AF 7)
s s ksi sy = 6.0 y = 28
47
s
y
s 11 75. = ksi
==
=2
FsπD4
N


4
1


=2
12 11 75. π D 4
D = 14.1 in
1
say D in
a D 97.0 in = =
=1
8
.0 8604 .0 8604 1  =
8

but a > D
1
say a in
b 48.0 in
=
1
.0 4302 say b in
1= 
8
=
2


1


1



=1
4
1

say c


=1in
81
d 05.2 in
=
say d =
.1 8257 2 in
1= 
8


=
1



c .1 030 in
=
1=
.0 9129
m 86.0 in
=
say m
.0 7607 in
1= 
8
7


8
1





1
n 71.1 in
=
1=
.1 5214
say n in

8
3
=1
4
=1
4
Dimension:
1
a in
68
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
=
b in 2
c =1in
d=2
in 7
m in
=
8
3
=1
4
1
=1
8
n in
D in
48. Give all the strength equations for the union of rods shown.
Problems 48
– 68.
Solution:
1


=2


F s π d Equation (1)
4
F s ( ad ) = s π Equation (2)
F s ( tc) s = 2 Equation (3)
69
SECTION 1– DESIGN FOR SIMPLE STRESSES
F s [ (D e)b] = s2 Equation (4) −
F
=
set
Equation
F = s(D − e)t Equation (6)
(5)



()
 2 2
1
F s π k e Equation (7)
=−
4
1
π Equation (8)
()()


 
22
F=sm−e−m−et4
F s ( ef ) s = 2 Equation (9)
70
SECTION 1– DESIGN FOR SIMPLE STRESSES

1
π Equation (10) 
F=se 

− et 2 4 
49-68. Design a union-of-rods joint similar to that shown for a reversing load and
material given in the accompanying table. The taper of cotter is to be ½ in. in 12
in. (see 172). (a) Using design stresses based on yield strengths determine all
dimensions to satisfy the necessary strength equations. (b) Modify dimensions as
necessary for good proportions, being careful not to weaken the joint. (c) Decide
upon tolerances and allowances for loose fits. (d) Sketch to scale each part of the
joint showing all dimensions needed for manufacture, with tolerances and
allowances.
Prob. No. Load, lb. AISI No., As Rolled
49 3000 1020
50 3500 1030
51 4000 1117
52 4500 1020
52 5000 1015
54 5500 1035
55 6000 1040
56 6500 1020
57 7000 1015
58 7500 1118
59 8000 1022
60 8500 1035
61 9000 1040
62 9500 1117
63 10,000 1035
64 10,500 1022
65 11,000 1137
66 11,500 1035
67 12,000 1045
68 12,500 1030
71
SECTION 1– DESIGN FOR SIMPLE STRESSES
Solution: (For Prob. 49 only)
(a) For AISI 1020, as rolled
s ksi y = 48
s s ( ) ksi ys = 6.0 y = 6.0 48 = 28 8.
For reversing load, N = 4 based on yield strength
48
s
y
s 12 = = ksi
=
s
N
s
sys
4
28 8.
N
4
2.7 = = = ksi
F = 3000 lb = 3
kips Equation (1)
=2
Fsπd4
=2
3 12 π d 4
d = .0 5642
in 9
say d in
Equation (2) F
s ( ad ) = s π
=
169

( ) 
a = .0 236
in
1
say a in
Equation (5)
F = set
3 =12et
et = 25.0
Equation (10)


1




1


=
16
3 2.7 π 

a
=


4
1
π


F = s e − et 2 4


3 12 2
πe
= − 25.0 4
e = .0 798 in
13
say e in


1
=
16
et = 25.0
72
SECTION 1– DESIGN FOR SIMPLE STRESSES

25.0
13
 = 16


t
t = ..0 308
in 5
say t in
=
Equation (6)
F = s(D − e)t
13
3 12 D
=−
165


16
D = .1 6125
in
5
say D in
Equation (4)
F s [ (D e)b] = s2
 
−  
=−b
8
 


=1


16


8
3 2.7 2 5
1
13


b = .0 say b in
256 in 1 16
4
=
Equation (7)


1

)
22

(
Fsπke
=−4
  
 

1


3 12 π k
=
k = .0 989 in
say k =1in
Equation (9)
F s ( ef ) s =
2
= 13
16 3 2.7 2
()f
f = .0 256 in
1
say f in
Equation (8)


1
π
2



2

 
13
4
 −  16



=
4


 2 2
()()
F=sm−e−m−et
4
  
  
 
  
2

 
=−
165


1
2
3 12
πmm


4
16


13
−−
  
 
13
16


73
SECTION 1– DESIGN FOR SIMPLE STRESSES
25.0 .0 7854 .0 5185 .0 3125 .0 2539 2
=m−−m+
2
m−m−=
.0 7854 .0 3125 .0 5146 0
2
m−m−=
.0 3979 .0 6552 0
m = .1 032 in
say m =1in
Equation (3)
F s ( tc) s = 2
=
165
c
( )  


3 2.7 2
c = .0 667 in
11
say c in
=
16
=
16 1
=
4
1
=
4
11
=
1
6
1
DIMENSIONS:
9
d in
a in
b in
c in
=
f in 4
13
e in
=
16 5
=
16
=1
8
=
16 1
=
4
3
=
4
11
=
16
t in
k =1in
5
D in
m =1in
(b) Modified dimensions 9
d in
a in
b in
c in
74
SECTION 1– DESIGN FOR SIMPLE STRESSES
1
=
f in 2
13
e in
=
16 5
=
16
=1
81
=1
4
t in
k =1in
5
D in
m in
(c) Tolerances and allowances
No fit, ± .0 010 in
d = .0 5625 ± .0 010 in
a = .0 250 ± .0 010 in
f = .0 500 ± .0 010 in
D = .1 625 ± .0 010 in
k = .1 000 ± .0 010 in
m = .1 250 ± .0 010 in
Fits, Table 3.1, loose-running fits, say RC 8
Female Male
.0 750
−+
.0 0035
b in
= b in
=
.0 0000
allowance = 0.0045 in
.0 0028
.0 6875
.0 6875
−−
−+
c in
= c in
=
.0 0000
allowance = 0.0035 in
.0 0035
.0 8125
.0 8125
−−
−+
= e in
=
e in
.0 0000
.0 0022
.0 3125
.0 3125
−−
−+
t in
= t in
=
.0 0000
allowance = 0.0030 in
.0 750
−−
.0 0045
.0 0065
.0 0035
.0 0051
.0 0045
allowance = 0.0045
.0 0065
.0 0030
.0 0040
75
SECTION 1– DESIGN FOR SIMPLE STRESSES
(d)
ROD
COTTER
76
SECTION 1– DESIGN FOR SIMPLE STRESSES
SOCKET
CHECK PROBLEMS
1
1
69. The connection shown has the following dimensions: d in
1
1 1 = , h in 5
D in
= , t in
4
1
=2, 2
= 1 , D in
= ; it supports a load of 15 kips.
Compute the tensile,
2
8
2
compressive, and shear stresses induced in the connection. What is the
corresponding design factor based on the yield strength if the rod and nut are
made of AISI C1045, as rolled, and the plate is structural steel (1020)?
77
SECTION 1– DESIGN FOR SIMPLE STRESSES
Problem 69.
Solution:
Tensile Stresses
15
F
(1) ksi s 12 22. 1 =
==
12
2
1


π
1
1
d
4
F


4
15


1
1


4
π
(2) ksi s 4.8 2 =
==
12
2
1
π
πD
1
4
Compressive Stress
F
(3)
s 78.4 3 =
4
2
15
=
ksi
()
=
1 22

 
1


1

π
2
DD
2

 
 


1
1
4
−
Shear 1
Stresse
s F
4
π
15
2
−
2
2
(4) ksi ss82.3 4=
π
Dt
π


==
2
2
(5) ksi ss09.5 5=
1
1
  
 2
F
15
==

2
π
π

Dh1
For AISI C1045, as rolled (rod and nut)
s ksi y 59
1=
s s ( ) ksi ys 6.0 y 6.0 59 35 4.
1= = =
For structural steel plate (1020)
s ksi y 48
2=
s s ( ) ksi ys 6.0 y 6.0 48 28 8.
1= = =
Solving for design factor
1
1


5
  
 8


78
SECTION 1– DESIGN FOR SIMPLE STRESSES
Ny
59
s
(1) 83.4
1
===1
12 22. 59
ss
1
Ny
(2) 95.6
1
===2
s
s
2
s
78.4
s
49.8
48
Ny
(3) 10 04.
2
===3
3
ys
28 8.
(4) 54.7
N
4= = = 2
s
s
4
ys
s
82.3
35 4.
(5) 96.6
N
5= = = 1
s
s
09.5
5
The corresponding design factor is N = 83.4
7
3
3
= , t in
= , b in
70. In the figure, let D in 4
= 3 , and let the load, which is applied 16 4
centrally so that it tends to pull the plates apart, be 15 kips. (a) Compute the
stresses in the various parts of the connection. (b) If the material is AISI C1020,
as rolled, what is the design factor of the connection based on yield strengths?
Problem 70.
Solution:
(a) Tensile stresses
15
F
s 11 43. 1 =
=
()
ksi
t b D −=
s 11 43. 2 =

34
F
16


7

  
  
−

434 4
3 

 ( ) 15
3
3
ksi
=
D
()tb =


−
16

Compressive bearing stress
−
2

7
3
3
4


2
  
 

3

4
F
15
4
  
 16
s 11 43. 3 =
==
3
ksi
7
4
Dt
4




79
SECTION 1– DESIGN FOR SIMPLE STRESSES
Shearing
stress F
=
15
ss24.4 4 =


()
4
π
π
1
()2


=


3


2
ksi
4
D
2
s
Ny
= or
s
N=
ys
4
2
(b) For AISI C1020, as rolled
s ksi y = 48
s s ksi ys = 6.0 y = 28 8.
Ny
s
=
Using s
s
ss
Ny
48
s
===
2.4
Using
s
N=
11.43
s
ys
s
s
N
s
=== s
24.4
28 8. 8.6
Therefore the design factor is N = 2.4
s
ys
15
9
3
1
71. For the connection shown, let a in
= , b in
16
3
15
16
= , c in
= , d in
4
=1 , 2
= , m n in
D in 4
annealed (see Fig. AF 1). 16
= = . The material is AISI C1040,
(a) For a load of 7500 lb., compute the various tensile, compressive, and shear
stresses. Determine the factor of safety based on (b) ultimate strength, (c) yield
strengths.
Problem 71.
Solution:
(a) Tensile stresses
80
SECTION 1– DESIGN FOR SIMPLE STRESSES
5.7
F
s 16 98. 1 =
12
ksi
==
2
1


3
4
π
D
4
F
π


4
5.7
s 89.8 2 =
=
()
ksi
2
bdc−=
2
16
 
 


−


9
1
1

5.7
2
4
F

3
s 10 67. 3 =
=
()
ksi


adc−=

15
−

16

 
  
1
1
3
2
4
Compressive Stresses (Bearing)
F
5.7
9
16
3
ksi
  
 4
s 89.8 4 =
2
bc
2
==




5.7
F
s 10 67. 5 =
= = ac 

15 16   
 3 4


16
  
 16
F
5.7



2
16


nb
For AISI C1040, annealed,Fig. AF 1
s ksi y = 47
s ksi
u = 79
s s ksi ys = 6.0 y = 28
  
 16


ksi
F
Shearing 5.7
Stresses
ss56.3 6=
==
15
4
ksi
9
4
mb


ss11.7 7=
==
15
2
9
ksi
s s ksi us u = 0.6 = 47.4
(b) Based on ultimate strength
79
s
Nu
===
65.4
1
s
16 98.
(c) Based on yield strength
47
s
Ny
===
77.2
s
1
16 98.
72. The upper head of a 60,000-lb. tensile-testing machine is supported by two steel rods,
one of which A is shown. These rods A are attached to the head B by split rings C.
The test specimen is attached to the upper head B so that the tensile force
81